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Hint: Just go on differentiating the both terms separately as it is given with respect to $x$ and the variable in question is $x$, you can directly differentiate by basic properties $d\left( a+b \right)=da+db$ . By this differentiate each term and then add them to get the final answer. Assume each term as a different variable and apply log to find the differentiation of that term significantly.

Complete step-by-step answer:

Given expression in the question for which we need differentiation

${{\left( \log x \right)}^{x}}+{{x}^{\log x}}$

We separate both the terms from the given expression above:

Let us assume that the term ${{\left( \log x \right)}^{x}}$ represented by “a”

Let us assume that the second term ${{x}^{\log x}}$ represented by “b”

Let us assume the total given expression represented by $''y''$

So, now we get that from above assumption we say:

$\begin{align}

& y={{\left( \log x \right)}^{x}}+{{x}^{\log x}}=a+b \\

& y=a+b \\

\end{align}$

By differentiating with respect to $x$ on both sides, we get

$\dfrac{dy}{dx}=\dfrac{da}{dx}+\dfrac{db}{dx}..........\left( i \right)$

From our first assumption we say that the value of “a” is:

$a={{\left( \log x \right)}^{x}}$

By applying the logarithm on both sides above, we get:

$\log a=x\log \left( \log x \right)$

Differentiating on both sides with respect to $x$ on both sides, we get:

$\dfrac{1}{a}\dfrac{da}{dx}=\log \left( \log x \right)+\dfrac{x}{\log x}\left( \dfrac{1}{x} \right)$

We use the formulas: $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x};d\left( uv \right)=udv+vdu$

From above, we get:

$\dfrac{da}{dx}=a\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)..........\left( ii \right)$

We know

\[b={{x}^{\log x}}\]

By applying logarithm on both sides of equation, we get:

$\log b={{\left( \log x \right)}^{2}}$ . We know $\dfrac{d}{dx}{{x}^{2}}=2x$

By differentiating with respect to $x$ on both sides, we get:

$\dfrac{1}{b}\dfrac{db}{dx}=\dfrac{2\log x}{x}$

By above we can say that value of the term $\dfrac{db}{dx}=2b\dfrac{\log x}{x}.........\left( iii \right)$

By equation (i) of differential, we get the value of $\dfrac{dy}{dx}:$

$\dfrac{dy}{dx}=\dfrac{da}{dx}+\dfrac{db}{dx}$

By substituting equation (ii) and equation (iii) above, we get

$\dfrac{dy}{dx}=a\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)+2b\dfrac{\log x}{x}$

By substituting a, b values back into equation, we solve it to:

$\dfrac{dy}{dx}={{\left( \log x \right)}^{x}}\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)+\dfrac{2{{x}^{\log x}}\log x}{x}$

Therefore, this is the final expression.

Note: i) While applying log to a, don’t forget to write $\log \left( \log x \right)$ some may confuse and write $\log x$ . But log should come twice.

(iI) In second term $\log b$ is directly ${{\left( \log x \right)}^{2}}$ you may write $\left( \log x \right)\left( \log x \right)$ for clarity.

Complete step-by-step answer:

Given expression in the question for which we need differentiation

${{\left( \log x \right)}^{x}}+{{x}^{\log x}}$

We separate both the terms from the given expression above:

Let us assume that the term ${{\left( \log x \right)}^{x}}$ represented by “a”

Let us assume that the second term ${{x}^{\log x}}$ represented by “b”

Let us assume the total given expression represented by $''y''$

So, now we get that from above assumption we say:

$\begin{align}

& y={{\left( \log x \right)}^{x}}+{{x}^{\log x}}=a+b \\

& y=a+b \\

\end{align}$

By differentiating with respect to $x$ on both sides, we get

$\dfrac{dy}{dx}=\dfrac{da}{dx}+\dfrac{db}{dx}..........\left( i \right)$

From our first assumption we say that the value of “a” is:

$a={{\left( \log x \right)}^{x}}$

By applying the logarithm on both sides above, we get:

$\log a=x\log \left( \log x \right)$

Differentiating on both sides with respect to $x$ on both sides, we get:

$\dfrac{1}{a}\dfrac{da}{dx}=\log \left( \log x \right)+\dfrac{x}{\log x}\left( \dfrac{1}{x} \right)$

We use the formulas: $\dfrac{d}{dx}\left( \log x \right)=\dfrac{1}{x};d\left( uv \right)=udv+vdu$

From above, we get:

$\dfrac{da}{dx}=a\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)..........\left( ii \right)$

We know

\[b={{x}^{\log x}}\]

By applying logarithm on both sides of equation, we get:

$\log b={{\left( \log x \right)}^{2}}$ . We know $\dfrac{d}{dx}{{x}^{2}}=2x$

By differentiating with respect to $x$ on both sides, we get:

$\dfrac{1}{b}\dfrac{db}{dx}=\dfrac{2\log x}{x}$

By above we can say that value of the term $\dfrac{db}{dx}=2b\dfrac{\log x}{x}.........\left( iii \right)$

By equation (i) of differential, we get the value of $\dfrac{dy}{dx}:$

$\dfrac{dy}{dx}=\dfrac{da}{dx}+\dfrac{db}{dx}$

By substituting equation (ii) and equation (iii) above, we get

$\dfrac{dy}{dx}=a\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)+2b\dfrac{\log x}{x}$

By substituting a, b values back into equation, we solve it to:

$\dfrac{dy}{dx}={{\left( \log x \right)}^{x}}\left( \log \left( \log x \right)+\dfrac{1}{\log x} \right)+\dfrac{2{{x}^{\log x}}\log x}{x}$

Therefore, this is the final expression.

Note: i) While applying log to a, don’t forget to write $\log \left( \log x \right)$ some may confuse and write $\log x$ . But log should come twice.

(iI) In second term $\log b$ is directly ${{\left( \log x \right)}^{2}}$ you may write $\left( \log x \right)\left( \log x \right)$ for clarity.

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